Photon Momentum

OpenStaxCollege; Saul Beceiro-Novo

Introduction to Quantum Physics

99 Photon Momentum

Relativistic Photon Momentum

The momentum carried by a photon is closely related to its energy. In special relativity, the total energy of a particle is related to its momentum and rest mass by

[latex]E^2=(pc)^2+(mc^2)^2.[/latex]

For a photon, the rest mass is zero:

[latex]m=0.[/latex]

However, the photon momentum is not zero. Substituting [latex]m=0[/latex] into the relativistic energy–momentum relationship gives

[latex]E^2=(pc)^2.[/latex]

Taking the positive square root gives

[latex]E=pc.[/latex]

Therefore, the momentum of a photon can also be written as

[latex]p=\frac{E}{c}.[/latex]

This expression applies specifically to photons and other massless particles. It shows that a photon with greater energy also carries greater momentum.

We can confirm that this relationship is consistent with the previously established expression for photon momentum. The energy of a photon is

[latex]E=\frac{hc}{\lambda}.[/latex]

Substituting this expression into [latex]p=E/c[/latex] gives

[latex]p= \frac{hc/\lambda}{c} = \frac{h}{\lambda}.[/latex]

Thus, the two expressions for photon momentum are equivalent:

[latex]p=\frac{E}{c}=\frac{h}{\lambda}.[/latex]

These relationships connect the particle and wave descriptions of electromagnetic radiation. The expression [latex]p=E/c[/latex] relates momentum to photon energy, while [latex]p=h/\lambda[/latex] relates momentum to wavelength.

Healthcare Connection: Photon Detectors

Most systems used to detect light rely on the particle-like interaction of individual photons with sensitive materials. Examples include the human eye, photographic film, photomultiplier tubes, charge-coupled devices (CCDs), and semiconductor detectors.

When a photon is absorbed, it transfers energy and momentum to an atom, molecule, or electron in the detector. This interaction may produce an electrical signal, a chemical change, or the emission of additional electrons. The individual signals are then amplified and combined to create a measurable result or an image.

Photon detectors are essential in biomedical imaging systems, including digital radiography, computed tomography, nuclear medicine, fluorescence microscopy, and positron emission tomography. Improving detector efficiency allows more of the available photons to contribute to the image, potentially improving image quality or reducing the radiation dose required from the patient.

Some highly sensitive detectors are cooled to reduce thermal noise. Thermal motion can produce electrical signals that resemble genuine photon detections, so reducing the detector temperature helps distinguish actual photons from background fluctuations.

Example 99.2: Photon Energy and Momentum

Problem

Show that [latex]p=E/c[/latex] gives the same momentum for the 500-nm photon considered in Example 99.1.

Strategy

Use the photon energy found in Example 99.1, convert it from electron volts to joules, and divide by the speed of light. The result can then be compared with the momentum calculated using [latex]p=h/\lambda[/latex].

Solution

The photon energy is

[latex]E=2.48\ \text{eV}.[/latex]

Converting this energy to joules gives

[latex]E= (2.48\ \text{eV}) \left( 1.602\times10^{-19}\ \frac{\text{J}}{\text{eV}} \right) = 3.97\times10^{-19}\ \text{J}.[/latex]

The photon momentum is therefore

[latex]p= \frac{E}{c} = \frac{3.97\times10^{-19}\ \text{J}} {3.00\times10^8\ \text{m/s}} = 1.33\times10^{-27}\ \text{kg}\cdot\text{m/s}.[/latex]

Discussion

This is the same momentum obtained in Example 99.1 using

[latex]p=\frac{h}{\lambda}.[/latex]

The agreement confirms that

[latex]p=\frac{E}{c} \qquad\text{and}\qquad p=\frac{h}{\lambda}[/latex]

are equivalent expressions for photon momentum.

This result also confirms that the relativistic energy–momentum relationship

[latex]E^2=(pc)^2+(mc^2)^2[/latex]

applies to photons as well as to particles with mass. A photon has zero rest mass, but it still possesses both energy and momentum.

Problem-Solving Suggestion

The following forms of Planck's constant are especially convenient when solving photon-energy and photon-momentum problems:

[latex]h=4.14\times10^{-15}\ \text{eV}\cdot\text{s}[/latex]

and

[latex]hc=1240\ \text{eV}\cdot\text{nm}.[/latex]

Use units that match the information provided in the problem. When wavelength is given in nanometers and energy is requested in electron volts, the value [latex]hc=1240\ \text{eV}\cdot\text{nm}[/latex] often simplifies the calculation.

Section Summary

Photons carry linear momentum even though they have zero rest mass.
The momentum of a photon is related to its wavelength by